The following solutions were… - NEET UG 2021 Chemistry PYQ

NEET UG 2021 — Chemistry Physical Chemistry

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The following solutions were prepared by dissolving $10g$ of glucose $({C}_{6}{H}_{12}{O}_{6})$ in $250\mathrm{mL}$ of water $({P}_{1}),10g$ of urea $({\mathrm{CH}}_{4}{N}_{2}O)$ in $250\mathrm{mL}$ of water $({P}_{2})$ and $10g$ of sucrose $({C}_{12}{H}_{22}{O}_{11})$ in $250\mathrm{mL}$ of water $({P}_{3}).$ The right option for the decreasing order of osmotic pressure of these solutions is:

Official previous-year question

Held on 30 Apr 2021 · Verified 9 Jul 2026.

Options

A
${P}_{2}>{P}_{1}>{P}_{3}$
B
${P}_{1}>{P}_{2}>{P}_{3}$
C
${P}_{2}>{P}_{3}>{P}_{1}$
D
${P}_{3}>{P}_{1}>{P}_{2}$

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Solution

Osmotic pressure $(\pi )=\mathrm{CRT}$

At same temperature, $\pi \propto C$

Solution: $(1)$ ${C}_{1}=\frac{10}{180}=\frac{10}{180}\times 4$

${C}_{1}=\frac{250}{1000}=\frac{40}{180}=\frac{2}{9}$

${C}_{1}=0.222M$

Solution: $(2)$ ${C}_{2}=\frac{10}{\frac{60}{\frac{250}{1000}}}=\frac{1}{6}\times 4=0.667M$

Solution: $(3)$ ${C}_{3}=\frac{10}{342}\times \frac{1000}{250}=0.1169M$

$\because {C}_{2}>{C}_{1}>{C}_{3}\therefore$ Hence, order: ${P}_{2}>{P}_{1}>{P}_{3}$

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Solution

Osmotic pressure $(\pi )=\mathrm{CRT}$

At same temperature, $\pi \propto C$

Solution: $(1)$ ${C}_{1}=\frac{10}{180}=\frac{10}{180}\times 4$

${C}_{1}=\frac{250}{1000}=\frac{40}{180}=\frac{2}{9}$

${C}_{1}=0.222M$

Solution: $(2)$ ${C}_{2}=\frac{10}{\frac{60}{\frac{250}{1000}}}=\frac{1}{6}\times 4=0.667M$

Solution: $(3)$ ${C}_{3}=\frac{10}{342}\times \frac{1000}{250}=0.1169M$

$\because {C}_{2}>{C}_{1}>{C}_{3}\therefore$ Hence, order: ${P}_{2}>{P}_{1}>{P}_{3}$

Options

Did you get this right?

Solution

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Organic Chemistry

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Options

Did you get this right?

Solution

Your note

NEET UG Physical Chemistry in other years

More NEET UG Physical Chemistry questions

Other Chemistry topics for NEET UG

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry